Does Fourier series of x^2 converge?

I'm trying to show that the Fourier series of [itex] f(x)=x^2[/itex] converges and I can't. Does anybody know if it actually does converge? (I'm assuming that [itex] f(x)=x^2[/itex] for [itex] x\in [-\pi,\pi][/itex]).
The Fourier Series itself is [itex] \displaystyle\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos nx [/itex]
I tried using Dirichlet's test but it wasn't working for me, though that may be because I'm doing something wrong.

I was just looking to see if it converged. So yes, you're absolutely right. I was going well off-beam with my attempt.

And about converging to [itex]x^2[/itex]; I guess it does so uniformly since the function is continuous on the circle and the Fourier series converges absolutely.

Thank you!

It's not because a function is continuous and because the Fourier series converges absolutely, that you can have uniform convergence (I think).
Here, I think you can infer uniform convergence from the Weierstrass M-test.

No I think it is. I quote Corollary 2.3 from "Fourier Analysis" by Stein and Shakarchi -

"Suppose that [itex]f [/itex] is a continuous function on the circle and that the Fourier series of [itex]f [/itex] is absolutely convergent, [itex]\sum_{n=-\infty}^\infty |\hat{f}(n)|<\infty. [/itex] Then, the Fourier series converges uniformly to [itex]f[/itex], that is
[itex]\displaystyle \lim_{N\to\infty}S_N(f)(\theta)=f(\theta) [/itex] uniformly in [itex]\theta. [/itex]